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# Hot water cools from 60degree celcius to 50 degree celcius in the first 10 min and to 42 degree celcius in the next 10 min .The temp of the surrounding. 12 years ago

average temp. in first case = (60+50) /2 = 55 C

average temp. in second case = (50+42) /2 = 51 C

as frm law of cooling

dT/dt = -k(T-To )  where To is temp. of surrounding

case 1 : dT = 10 C and dt = 10 mins

so ,  1=-k(55-To )         ....(1)

case 2 : dT = 8 C and dt = 10 mins

so ,  0.8=-k(51-To )       .........(2)

From 1 and 2 , we have

Temp of surr. as 35 C

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Regards,
Naga Ramesh
IIT Kgp - 2005 batch

5 years ago
Excuse sir I need to correct u in the second step ie., (50+42)/2=46
So the room temp would be 10C
dT/dt=0.8=-k (46-t)
dT2/dt2 =k (55-t)
Dividing them both we get
1/0.8=(55-t)/(46-t)
Therfore t=10
Answer by Aswath S, AECS PU College, Bangalore Fiitjee
Future IIT ian     ha ha ha Kushagra Madhukar
one year ago
Dear student,

Average temp. in first case = (60+50) /2 = 55 C
Average temp. in second case = (50 + 42) / 2 = 46 C
from law of cooling
dT/dt = – k(T – To )  where To is temp. of surrounding

Case 1 : dT = 10 C and dt = 10 mins
so ,  1 = – k(55 – To)
Case 2 : dT/dt = 0.8 = – k (46 – To)
0.8 = k(55 – To)
Dividing them both we get
1/0.8 = (55 – To) / (46 – To)
On solving, we get, To = 10oC

Hope it helps.
Thanks and regards,
Kushagra